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Attempting to find an inductive argument for the Borsuk-Ulam theorem led me to another question, which I found interesting in its own right but am stuck on.

Let $g:S^n\rightarrow \mathbb{R}$ be a $C^{\infty}$ mapping such that $\forall x \in S^n g(x)=-g(-x)$, and $0$ is a regular value of $g$.

It follows easily that $A=g^{-1}(0)$ is a closed $n-1$ manifold, and that there is necessarily a connected component $a\in A$ such that $a=-a$ (this comes from the fact that all antipode pairs in $S^n - A$ are path-disconnected, and thus that $S^n - A$ has an even number of connected components.)

However, is there anything more that can be said about the particular connected component of $A$ which is fixed under antipode transposition? I've been attempting to look up stuff about regular values and their preimages online, but without much luck as I'm not really sure what I should be looking for. In the case of $S^2$, this component obviously must be a circle, and in $S^3$ I know it can be a 2-sphere, and think I have constructed a function $g$ such that it is also a genus 2 surface, but an unsure what can be said in general. Any pointers at all would be helpful really, this is just something I've been working on for fun, but unfortunately have not yet taken enough math yet to know where to look next.

Thanks!

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Not about the particular (antipode stable) component of $A$ but about $A$ itself: It's the boundary of a smooth manifold, which actually says something about it. All my thoughts lead to cobordism theory — you could have a look at Milnors Topology from the Differentiable Viewpoint (about regular values and (framed) cobordism). I wonder if there is a criterion to check whether $A$ is connected. –  Ben A. Apr 3 '12 at 15:59

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